Solution to problem set 3

1. $SU(N)$ Anderson Impurity model

   We consider impurity electrons on a single site hybridizing with a bath of conduction electrons. The electrons carry $SU(N)$ spin. We consider the limiting case where the on-site Coulomb repulsion ($U$) at the impurity site is infinitely large. Thus, at most a single electron can occupy the impurity site (multiple occupancy is not allowed). In this limit, the allowed states of the impurity site are $\vert 0 \rangle$ (i.e., no electron), and $\vert m \rangle ,$ where $m= (1, \cdots ,N)$ (i.e., a single electron with $SU(N)$ spin $m$). The Hubbard operators are defined as $X_{pq} = \vert p \rangle \langle q \vert$, where $p,q = (0,1, \cdots, N)$. In this restricted subspace creation operator for an impurity electron $f^{\dagger}_n \equiv \vert n \rangle \langle 0 \vert = X_{n0}$. Similarly, $f_n \equiv X_{0n}$. The completeness relation in this subspace (of the impurity level) is given by $X_{00} + \sum_n X_{nn} = 1$. Note, that the Hubbard operators satisfy the relation $[X_{pq} , X_{rs} ]_{\pm} = \delta _{qr} X_{ps} \pm \delta _{ps} X_{rq}$. This is unlike standard commutation or anticommutation of bosons and fermions. The Hamiltonian for the system is given by
   

   $${\mathcal H}= {\mathcal H}_b + {\mathcal H}_f + {\mathcal H}_I,$$
   
   
   where ${\mathcal H}_b = \sum_{{\bf k}, n} \epsilon _k c^{\dagger}_{{\bf k}n} c_{{\bf k}n}$ describes the bath of conduction electrons, ${\mathcal H}_f = \epsilon _f \sum_m f^{\dagger}_m f_m = \epsilon _f \sum_m X_{mm}$ describes the impurity electrons, and ${\mathcal H}_I = \sum_{{\bf k}m} ( V_{{\bf k}} c^{\dagger}_{{\bf k}m} X_{0 m} + V^{\ast}_{{\bf k}} X_{m 0} c_{{\bf k}m})$ is the hybridization between the impurity electrons and the conduction electrons.

   We consider the retarded Green's function $\langle \langle f_n f^{\dagger}_n \rangle \rangle (\omega ) = \langle \langle X_{0n} X_{n 0} \rangle \rangle (\omega )$. The corresponding spectral function is given by
   

   $$\rho_{f f^{\dagger}}(\omega)= - \frac{{\rm Im}}{ \pi} G_{f f^{\dagger}} ( \omega + i \delta ).$$
   
   

   The first moment is given by $\langle \{ X_{0n} , X_{n0} \} \rangle = \langle X_{00} + X_{nn} \rangle$. We can write $\langle X_{00} \rangle = 1 - \langle n_f \rangle$, where $\langle n_f \rangle = \sum_n \langle X_{nn} \rangle$ is the expectation of electron occupancy at the impurity site. Also, $\langle X_{nn} \rangle = \langle n_f \rangle /N$. Thus,
   

   $$\langle \{ X_{0n} , X_{n0} \} \rangle = 1 - \left( \frac{N-1}{N} \right) \langle n_f \rangle .$$ (1)

   
   
   Since,
   
   $$\int_{- \infty}^{\infty} d \omega \rho_{f f^{\dagger}}(\omega)= \langle \{ X_{0n} , X_{n0} \} \rangle ,$$
   
   
   the first moment gives the total spectral weight of the f-band.

   For the calculation of the second moment we get
   

   $$[ {\mathcal H}, X_{0n} ]= - \epsilon _f X_{0n} - \sum_{{\bf k... ...00} - \sum_{{\bf k}m } V^{\ast}_{{\bf k}} c_{{\bf k}m} X_{mn}.$$
   
   
   Then,
   

   $$\langle \{ [ {\mathcal H}, X_{0n} ] , X_{n0} \} \rangle = - ... ...eq n} V^{\ast}_{{\bf k}} \langle c_{{\bf k}m} X_{m0} \rangle .$$ (2)

   
   
   Since,
   

   $$\int_{- \infty}^{\infty} \omega \rho_{f f^{\dagger}}(\omega)... ... = - \langle \{ [ {\mathcal H}, X_{0n} ] , X_{n0} \} \rangle ,$$ (3)

   
   
   the centre of the f-band is given by
   
   $$\frac{\int_{- \infty}^{\infty} \omega \rho_{f f^{\dagger}}(\... ...eq n} V^{\ast}_{{\bf k}} \langle c_{{\bf k}m} X_{m0} \rangle .$$
   
   
   Presumably (since we cannot conclusively say anything about the line shape) the spectral function is peaked at this energy. The second term is the shift in the energy of the f-level due to hybridization.

   For the third moment we get
   

   $$\langle \{ [ {\mathcal H}, X_{0n} ], [ {\mathcal H}, X_{n0} ] \} \rangle = \langle \{ - \epsilon _f X_{0n} - \sum_{vk} V_{{\bf k}}^{\ast} c_... ...\sum_{{\bf k}m} V_{{\bf k}}^{\ast} c_{{\bf k}m} X_{mn} \ , \ \epsilon _f X_{n0}$$

   $$+ \sum_{{\bf k}} V_{{\bf k}} c^{\dagger}_{{\bf k}n} X_{00} + \sum_{{\bf k}m} V_{{\bf k}} c^{\dagger}_{{\bf k}m} X_{nm} \} \rangle$$

   $$= - \epsilon _f^2 \langle \{ X_{0n} , X_{n0} \} \rangle - \epsilon ... ...m_{{\bf k}, m \neq n} V_{{\bf k}} \langle c^{\dagger}_{{\bf k}m} X_{0m} \rangle$$

   $$- \sum_{{\bf k}} \vert V_{{\bf k}} \vert^2 + \sum_{{\bf k}_1 {\bf... ...k}_2} \langle c^{\dagger}_{{\bf k}_2 m_2} c_{{\bf k}_1 m_1} X_{m_1 m_2} \rangle$$

   $$- \sum_{{\bf k}_1 {\bf k}_2, m} V_{{\bf k}_1}^{\ast} V_{{\bf k}_2} \langle c^{\dagger}_{{\bf k}_2 m} c_{{\bf k}_1 m} X_{nn} \rangle .$$ (4)

   
   
   We can show that
   

   $$\langle \{ [ {\mathcal H}, X_{0n} ] , X_{n0} \} \rangle = \frac{1}{Z} \sum_{nm} \left( e^{- \beta E_n} + e^{- \beta E_m} \r... ...) \langle n \vert X_{0n} \vert m \rangle \langle m \vert X_{n0} \vert n \rangle$$

   $$= - \langle \{ X_{0n} , [ {\mathcal H}, X_{n0} ] \} \rangle .$$ (5)

   
   
   This is due to time translation invariance of the correlation function in equilibrium. Using this result we get,
   

   $$- \sum_{{\bf k}, m \neq n} V_{{\bf k}}^{\ast} \langle c_{{\b... ...n} V_{{\bf k}} \langle c^{\dagger}_{{\bf k}m} X_{0m} \rangle .$$ (6)

   
   
   The $SU(N)$ spin at the impurity site can be written as ${\bf S}= \sum_{m_1 m_2} \Gamma ^M_{m_1 m_2} X_{m_1 m_2}$, where $\Gamma ^M$ are the generators of $SU(N)$ group, with $M = (1, \cdots ,(N^2-1))$. The spin of the conduction electrons in the momentum basis is similarly given by ${\bf s}_{{\bf k}_2 {\bf k}_1} = \sum_{m_1 m_2} c^{\dagger}_{{\bf k}_2 m_1} \Gamma ^M_{m_1 m_2} c_{{\bf k}_1 m_2}$. The generators of the group satisfy the completeness relation $\sum_M \Gamma ^M_{m_1 m_2} \Gamma ^M_{m_3 m_4} = \delta _{m_1 m_4} \delta _{m_2 m_3} - \frac{1}{N} \delta _{m_1 m_2} \delta _{m_3 m_4}$. Using this we can show that
   

   $$\sum_{{\bf k}_1 {\bf k}_2} V_{{\bf k}_1}^{\ast} V_{{\bf k}_2... ...\bf k}_2 m} c_{{\bf k}_1 m} X_{m^{\prime}m^{\prime}} \rangle .$$ (7)

   
   
   Thus, we can write
   

   $$\langle \{ [ {\mathcal H}, X_{0n} ], [ {\mathcal H}, X_{n0} ] \} \rangle = - \epsilon _f^2 \langle \{ X_{0n} , X_{n0} \} \rangle - 2 \epsilo... ... \langle c_{{\bf k}m} X_{m0} \rangle - \sum_{{\bf k}} \vert V_{{\bf k}} \vert^2$$

   $$+ \sum_{{\bf k}_1 {\bf k}_2} V_{{\bf k}_1}^{\ast} V_{{\bf k}_2} \langle {\bf s}_{{\bf k}_2 {\bf k}_1} \cdot {\bf S}\rangle .$$ (8)

   
   
   We identify the last term in the above equation as a spin-spin correlation function. The sum rule for the third moment is given by
   

   $$\int_{- \infty}^{\infty} \omega ^2 \rho_{f f^{\dagger}}(\ome... ...{\mathcal H}, X_{0n} ] , [ {\mathcal H}, X_{n0} ] \} \rangle .$$ (9)

   
   
   The width of the impurity band (standard deviation of a distribution, in general) is given by
   

   $${ \frac{\int_{- \infty}^{\infty} \omega ^2 \rho_{f f^{\dagger}}(\... ...} {\int_{- \infty}^{\infty} d \omega \rho_{f f^{\dagger}}(\omega)} \right)^2 =}$$

   $$- \frac{1}{\langle \{ X_{0n} , X_{n0} \} \rangle }\sum_{{\bf k}_1... ... k}, m \neq n} V^{\ast}_{{\bf k}} \langle c_{{\bf k}m} X_{m0} \rangle \right)^2$$ (10)

   
   
   Originally the f-level was a delta function peak at $\epsilon _f$. Due to hybridization there is a shift in the energy level, and also a broadening of the band.

2. f-sum rule

   The retarded current-current correlation function is defined by
   

   $$\Pi_{jj}^R (\omega ) = \langle \langle j,j \rangle \rangle (... ...\rangle \right\vert ^2 \frac{1}{\omega + E_n - E_m + i \eta }.$$ (11)

   
   
   The corresponding spectral function is
   

   $$\rho_{jj} (\omega ) = - \frac{1}{\pi} {\rm Im} \Pi_{jj}^R (\omega )$$

   $$= \frac{1}{Z} \sum_{nm} \left( e^{- \beta E_n} - e^{- \beta E_m} \r... ...ert \langle n \vert j \vert m \rangle \right\vert^2 \delta (\omega + E_n - E_m)$$

   $$= \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ j(t) , j] \rangle .$$ (12)

   
   
   In terms of the spectral function, the real part of the conductivity (which is what enters the f-sum rule) is given by,
   

   $$\sigma (\omega ) = - \frac{1}{V} \frac{1}{\omega } {\rm Im} \Pi_{jj}^R (\omega ) = \frac{\pi}{V} \rho_{jj} (\omega ).$$ (13)

   
   
   We define the polarization operator $P$ such that $\frac{\partial P}{\partial t} = j$. Then,
   

   $$\rho_{jj} (\omega ) = \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ \frac{\partial P}{\partial t}, j ] \rangle$$

   $$= - i \omega \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ P(t), j ] \rangle$$

   $$= -i \omega \rho_{Pj} (\omega ),$$ (14)

   
   
   where
   

   $$\rho_{Pj} (\omega ) = \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ P(t), j ] \rangle .$$ (15)

   
   
   Then,
   

   $$\int_{- \infty}^{\infty} d \omega \sigma (\omega ) = - \frac{\pi i}{V} \int_{- \infty} ^{\infty} d \omega \rho_{Pj} (\omega )$$

   $$= - \frac{\pi i}{V} \langle [P(0), j ] \rangle .$$ (16)

   
   
   The current operator is given by $j = \sum_{i = 1}^N (e p_i)/m$, where $p_i$ is the momentum of the i-th particle. The polarization operator is given by $P = \sum_{i = 1}^N e r_i$, where $r_i$ is the position of the i-th particle. Since,
   
   $$[P, j]= \frac{e^2}{m} \sum_{ij} [ r_i , p_j] = \frac{N e^2 i}{m},$$
   
   
   (in units where $\hbar = 1$), we finally get the f-sum rule as
   

   $$\int_{- \infty}^{\infty} \frac{d \omega } {\pi} \sigma (\omega ) = \frac{n e^2}{m},$$ (17)

   
   
   where $n = N/V$ is the density of particles.